Prove that $\vdash p \lor \lnot p$ is true using natural deduction

I would like not to use de morgan law, as you would need to include that as a premise. I was thinking of this proof.

1. • $\lnot (p \lor \lnot p) \quad \quad \quad (H)$
2. • • $p \quad \quad \quad \quad \quad (H)$
3. • • $p \lor \lnot p \quad \quad \; \;(\lor \text{I} 2)$
4. • • $ \bot \quad \quad \quad \quad \;\;(\lnot \text{E}1,3)$
5. • $\lnot p \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \; (\lnot \text{I}2 - 4)$
6. • $p \lor \lnot p \;\;\;\;\;\;\;\;\; \;\;\;\;\;\,(\lor \text{I}5)$
7. • $\bot \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,(\lnot \text{E}1,6)$
8. $p \lor \lnot p \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\;(\bot \text{E}1-7)$

The "standard" Natural Deduction proof of it is :

1) $\lnot (p \lor \lnot p)$ --- assumed [a]

2) $p$ --- assumed [b]

3) $p \lor \lnot p$ --- 2) : $\lor$I

4) $\bot$ --- 1), 3) : $\rightarrow$E

5) $\lnot p$ --- 2), 4) : $\rightarrow$I, discharging [b]

6) $p \lor \lnot p$ --- 5) : $\lor$I

7) $\bot$ --- 1), 6) : $\rightarrow$E

8) $p \lor \lnot p$ --- 1), RAA, discharging [a]

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As expected, we need the classical rule for indirect proof (RAA, which is equivalent to Double Negation) to prove it, because Excluded Middle : $p \lor \lnot p$ is not intuitionistically valid.